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It takes printer A 4 more minutes than printer B to print 40 [#permalink]
 03 Aug 2010, 12:13
Question Stats:
51% (04:04) correct
48% (03:00) wrong based on 2 sessions
It takes printer A 4 more minutes than printer B to print 40 pages. Working together, the two printers can print 50 pages in 6 minutes. How long will it take printer A to print 80 pages? A. 12 B. 18 C. 20 D. 24 E. 30
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where did you get the source of this problem? I got something complete different. So ok. A => B+4min = 40 pages B=> Bmin = 40 pages (B+4 + B)6 = 50 pages (2B + 4)6 = 50 12B + 24 = 50 12B = 26 B = 26/12 = 13/6 so.... A = B + 4 A = (13/6) + 4 (13/6 + 6/6 + 6/6 + 6/6 + 6/6)t = 80 (37/6)t = 80 t = 12.97.... someone care to explain this?
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bhandariavi wrote: It takes printer A 4 more minutes than printer B to print 40 pages. Working together, the two printers can print 50 pages in 6 minutes. How long will it take printer A to print 80 pages?
a. 12
b. 18
c. 20
d. 24
e. 30 Let the time needed to print 40 pages for printer A be a minutes, so for printer B it would be a-4 minutes. The rate of A would be rate=\frac{job}{time}=\frac{40}{a} pages per minute and the rate of B rate=\frac{job}{time}=\frac{40}{a-4} pages per minute. Their combined rate would be \frac{40}{a}+\frac{40}{a-4} pages per minute. Also as "two printers can print 50 pages in 6 minutes" then their combined rate is rate=\frac{job}{time}=\frac{50}{6}, so \frac{40}{a}+\frac{40}{a-4}=\frac{50}{6}. \frac{40}{a}+\frac{40}{a-4}=\frac{50}{6} --> \frac{1}{a}+\frac{1}{a-4}=\frac{5}{24}. At this point we can either try to substitute the values from answer choices or solve quadratic equation. Remember as we are asked to find time needed for printer A to print 80 pages, then the answer would be 2a (as a is the time needed to print 40 pages). Answer D works: 2a=24 --> a=12 --> \frac{1}{12}+\frac{1}{8}=\frac{5}{24}. Answer: D.
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I got this question - unanswered somewhere around page 19-20 in PS Discussion section. Is this a real GMAT type question?
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Solving work-rate equation (GMAT Club Test M17#20) [#permalink]
19 Oct 2010, 15:10
It takes printer A 4 more minutes than printer B to print 40 pages. If working together, the two printers can print 50 pages in 6 minutes, how long will it take printer A to print 80 pages? (C) 2008 GMAT Club - m17#20 * 12 * 18 * 20 * 24 * 30 I can get up to the part where: \frac{1}{A} + \frac{1}{A - 4} = \frac{5}{24}How do I solve this equation?? I wind up with some pretty complicated squared equation... Thanks! Solution: Denote A as the time it takes printer A to print 40 pages. Because working together the two printers can print 50 pages in 6 minutes, they can print 40 pages in \frac{40}{50}*6 = \frac{24}{5} minutes. Now, it follows from the stem that \frac{1}{A} + \frac{1}{B} = \frac{1}{A} + \frac{1}{A - 4} = \frac{5}{24} . The only acceptable solution to this equation is 12. So, printer A will print 40 pages in 12 minutes; therefore, it will print 80 pages in 24 minutes. |
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Re: Solving work-rate equation (GMAT Club Test M17#20) [#permalink]
19 Oct 2010, 15:13
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Re: Solving work-rate equation (GMAT Club Test M17#20) [#permalink]
19 Oct 2010, 15:16
Bunuel wrote: Merging similar topics.
As for your question from this point I recommend substitution from answer choices rather then solving quadratic equation. Ok, thanks. So I would be trying answer choices divided by 2, right? (since answers give A work for 80 pages, and equation is set-up based on 40 pages)
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Re: Solving work-rate equation (GMAT Club Test M17#20) [#permalink]
19 Oct 2010, 15:19
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B = x minutes for 40 pages A = x + 4 minutes for 40 pages A does in one in minute of 40 pages = 40/x+4 B does in one in minute of 40 pages = 40/x so, 6[(40/x+4)+40/x]= 50 5x^2 - 28x - 96 = 0 (x - 8)(x+12) = 0 x = 8 B = 8 minutes for 40 pages A = 12 minutes for 40 pages so, A needs 24 minutes for 80 pages.
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Got the quadratic soon but took ages after that to solve the rest of the problem. The substitution, I guess, would take up a lot of time. How to solve within 2 mins?
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Baten80 wrote: B = x minutes for 40 pages
A = x + 4 minutes for 40 pages
A does in one in minute of 40 pages = 40/x+4
B does in one in minute of 40 pages = 40/x
so, 6[(40/x+4)+40/x]= 50
5x^2 - 28x - 96 = 0
(x - 8)(x+12) = 0
x = 8
B = 8 minutes for 40 pages
A = 12 minutes for 40 pages
so, A needs 24 minutes for 80 pages. Can you explain this? 5x^2 - 28x - 96 = 0 (x - 8)(x+12) = 0 is there any faster way to solve such kind of quadratic expressions where "a" is different than 1??? Thanks
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Saurajm wrote: Baten80 wrote: B = x minutes for 40 pages
A = x + 4 minutes for 40 pages
A does in one in minute of 40 pages = 40/x+4
B does in one in minute of 40 pages = 40/x
so, 6[(40/x+4)+40/x]= 50
5x^2 - 28x - 96 = 0
(x - 8)(x+12) = 0
x = 8
B = 8 minutes for 40 pages
A = 12 minutes for 40 pages
so, A needs 24 minutes for 80 pages. Can you explain this? 5x^2 - 28x - 96 = 0 (x - 8)(x+12) = 0 is there any faster way to solve such kind of quadratic expressions where "a" is different than 1??? Thanks Check this: Factoring Quadratics: http://www.purplemath.com/modules/factquad.htmSolving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htmThough it's better to substitute the values rather than get the quadratics and then factor it (refer to my podt above to see how it can be done). Hope it helps.
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Re: It takes printer A 4 more minutes than printer B to print 40 [#permalink]
12 Feb 2012, 02:08
Dear Surajm
For this equation : 5x^2 - 28x - 96 = 0
The central term (-28) should be written is such a way which is combination of (5 * 96) [First term multiplied by last term] .
96*5 Can be factored as 8*2*6*5 (Prime factor will be 2*2*2*2*3*2*5)
Now -28 can be written as (-40 + 12)
-40 came from 8*5 from the bold factor above
12 came from 6*2 from the bold factor above
Hence
5x^2 - 28x - 96 = 0
5x^2 - 40x + 12x - 96 =0
5x(x-8) +12 (x-8) = 0
(5x +12)(x-8) = 0
==> x = 8 (Other value of x will be negative, hence avoid it)
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siddharthmuzumdar wrote: Got the quadratic soon but took ages after that to solve the rest of the problem. The substitution, I guess, would take up a lot of time. How to solve within 2 mins? Do not go up to the quadratic to solve it. The substitution will take a long time. Get your basic equation and then substitute. Let me show you what I mean. Let me make the work same. We need time taken by printer A to print 80 pages. (Let's say this is 'a' mins) We know A takes 4 more mins for 40 pages so it will take 8 more minutes for 80 pages. Together, they print 50 pages in 6 mins so they will print 80 pages in 6*50/80 = 48/5 mins Now, make your sum of rates equation: 1/a + 1/(a - 8) = 5/48 Now look at the options and substitute here. First check out the straight forward options. Say, a = 12 1/12 + 1/4 = 4/12 Nope I will not try 18 and 20 because (18, 10) and (20, 12) doesn't give me 48, the denominator on right hand side. I will try 24 instead. 1/24 + 1/16 = 5/48 Yes. Answer is 24.
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Re: It takes printer A 4 more minutes than printer B to print 40 [#permalink]
14 Nov 2012, 00:46
Rate of Printer A + Rate of Printer B = Rate Together
\frac{40}{a}+\frac{40}{a-4}=\frac{50}{6}
Simplify further:
\frac{1}{a}+\frac{1}{(a-4)}=\frac{5}{24}
The answer choices are equal to 2a = n.
Test all the answer choices and you will see D as the only one fulfilling the rate equation above.
D. 2a = 24 Thus a = 12
\frac{1}{12}+\frac{1}{(12-4)}=\frac{1}{12}+\frac{1}{8}=\frac{5}{24}
Answer: D
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Re: It takes printer A 4 more minutes than printer B to print 40
[#permalink]
14 Nov 2012, 00:46
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